A passenger train is traveling at 30 m/s when the engineer sees a freight train

Question

A passenger train is traveling at 30 m/s when the engineer sees a freight train 384 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 5.9 m/s.
(a) If the reaction time of the engineer is 0.42 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided?
(b) If the engineer's reaction time is 0.86 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide?

Explanation / Answer

a) V1 = 30 m/s v2 = 5.9 m/s x = x0 - (v1 + v2)*(t + 0.42) + 0.5*a*t^2 0 = 384 - 35.9*(t + 0.42) + 0.5*a*t^2 v = (v1 + v2) + a*t 0 = 35.9 + a*t solve for t = -35.9/a substitute t in the first eq 0 = 384 - 35.9*(-35.9/a + 0.42) + 0.5*a*(-35.9/a)^2 0 = 384 - 1288.81/a - 15 + 644.4/a 369 = 644.41/a ... so a = 644.41/369 = 1.74 m/sec^2 b) x = x0 - (v1 + v2)*(t + 0.86) + 0.5*a*t^2 0 = 384 - 35.9*(t + 0.86) + 0.5*1.74*t^2 0 = 353.12 - 35.9*t + 0.87*t^2 quadratic t = (-b+sqrt(b^2-4ac))/2a or (-b-sqrt(b^2-4ac))/2a t = 16.12 or 25 sec v = (v1 + v2) + a*t v = 35.9 + 1.74*16.12 = 63.94 m/sec or v = 35.9 + 1.74*25 = 79.4 m/sec

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